AN INTEGRAL UNIVALENT OPERATOR
D. BREAZ and N. BREAZ
Abstract. In this paper we consider the class of univalent functions defined by the condition˛˛
˛
z2f′(z)
f2(z) −1 ˛ ˛
˛<1, z∈U,wheref(z) =z+a2z
2+. . . ,is an analytic
function in the unit discU ={z∈C:|z|<1}.We present univalence conditions for the operator
Gα,n(z) =
0
@(n(α−1) + 1)
z
Z
0
g1α−1(t). . . gnα−1(t) dt
1
A 1
n(α−1)+1 .
1. Introduction
Let U be the unit disc, U = {z∈C:|z|<1}. Denote by H(U) the class of
holomorphic functions onU and consider the set of analytic functions
An =f ∈H(U) :f(z) =z+an+1zn+1+. . .
(1)
Ifn = 1 then A1 =A. Let Hu(U) be the class of univalent functions on the
unit discU andSthe class of regular and univalent functionsf(z) =z+a2z2+. . . inU,which satisfy the conditionf(0) =f′(0)−1 = 0.
In their paper [3], Ozaki and Nunokawa proved the following results:
Theorem 1. If we assumethat g∈Asatisfies the condition
z2f′(z) f2(z) −1
<1, z∈U (2)
thenf is univalent inU.
Lemma 1 (The Schwartz Lemma). Let the analytic function g be regular in the unit discU andg(0) = 0. If|g(z)| ≤1,∀z∈U, then
|g(z)| ≤ |z|, ∀z∈U (3)
and equality holds only if g(z) =εz, where |ε|= 1.
Received October 18, 2004.
2000Mathematics Subject Classification. Primary 30C45.
Theorem 2. Let α be a complex number with Reα > 0, and let f = z + a2z2+ . . . be a regular function onU. If
1− |z|2Reα Reα
zf′′(z)
f′(z)
≤1, ∀z∈U (4)
then for any complex numberβ withReβ≥Reαthe function
Fβ(z) =
β z Z
0
tβ−1f′(t) dt
1
β
=z+. . . (5)
is regular and univalent inU.
Theorem 3. Assume thatg∈Asatisfies condition (2), and let αbe a complex number with
|α−1| ≤ Reα 3 . (6)
If
|g(z)| ≤1, ∀z∈U (7)
then the function
Gα(z) =
α z Z
0
g(α−1)(t) dt
1
α
(8)
is of classS.
2. Main results
Theorem 4. Let gi ∈A,∀i= 1, . . . , n,n∈N∗, satisfy the properties
z2g′i(z)
g2
i (z)
−1
<1, ∀z∈U, ∀i= 1, . . . , n (9)
andα∈C, with
|α−1| ≤ Reα 3n . (10)
If|gi(z)| ≤1,∀z∈U,∀i= 1, . . . , n, then the function
Gα,n(z) =
(n(α−1) + 1) z Z
0
gα1−1(t). . . gnα−1(t) dt
1
n(α−1)+1 (11)
Proof. From (11) we have:
Gα,n(z) = ( (n(α−1) + 1) z Z
0
tn(α−1)
g1(t) t
α−1
·. . .
·
gn(t)
t
α−1
dt
!n(α−1)+11 (12)
We consider the function
f(z) =
z Z
0
g1(t) t
α−1
. . .
gn(t)
t
α−1
dt. (13)
The functionf is regular inU, and from (13) we obtain
f′(z) =
g1(z) z
α−1
. . .
gn(z)
z
α−1
(14)
and
f′′(z) = (α−1)
g1(z) z
α−2
zg1′(z)−g1(z) z2
g2(z) z
α−1
. . .
gn(z)
z
α−1
+. . .
+
g1(z) z
α−1
. . .
gn−1(z) z
α−1
(α−1)
gn(z)
z
α−2
zgn′ (z)−gn(z)
z2 .
(15)
Next we calculate the expresion zf′′
f′ .
zf′′(z)
f′(z) =
z
(α−1)g1(z)
z
α−2zg′
1(z)−g1(z)
z2 g
2(z)
z α−1
. . .gn(z)
z
α−1
g 1(z)
z α−1
. . .gn(z)
z
α−1 +. . .
+ z
g
1(z)
z α−1
. . .gn−1(z)
z
α−1
(α−1)gn(z)
z
α−2zg′
n(z)−gn(z)
z2
g 1(z)
z α−1
. . .gn(z)
z α−1
(16)
= (α−1)zg
′
1(z)−1
g1(z) +· · ·+ (α−1)
zg′n(z)−1
gn(z)
.
The modulus
zf′′
f′
can then be evaluated as
zf′′(z) f′(z)
=
(α−1)zg
′
1(z)−1
g1(z) +· · ·+ (α−1)
zg′n(z)−1
gn(z) ≤
(α−1)zg
′
1(z)−1 g1(z) +· · ·+
(α−1)zg
′
n(z)−1
gn(z)
=|α−1|
zg′
1(z)−1 g1(z)
+· · ·+|α−1|
zg′
n(z)−1
gn(z) . (18)
By multiplying the first and the last terms of (18) with 1−|Rez|2Reα α >0,we obtain
1− |z|2Reα Reα
zf′′(z)
f′(z)
≤ 1− |z| 2Reα
Reα |α−1|
zg′ 1(z) g1(z) + 1 +. . .
+1− |z| 2Reα
Reα |α−1|
zg′ n(z)
gn(z) + 1
≤ |α−1|1− |z| 2Reα Reα
z2g′
1(z) g2 1(z) |g1(z)| |z| + 1
+. . .
+|α−1|1− |z| 2Reα Reα
z2g′ n(z) g2 n(z)
|gn(z)|
|z| + 1
. (19)
By applying the Schwartz Lemma and using (19), we obtain
1− |z|2Reα Reα
zf′′(z) f′(z)
≤ |α−1|1− |z| 2Reα Reα
z2g1′ (z) g2 1(z) −1 + 2 +. . .
+|α−1|1− |z| 2Reα Reα
z2gn′ (z)
g2 n(z) −1 + 2 . (20)
Sincegi satisfies the condition (2)∀i= 1, . . . , n, then from (20) we obtain:
1− |z|2Reα Reα
zf′′(z) f′(z)
≤3|α−1|1− |z| 2Reα
Reα +· · ·+ 3|α−1|
1− |z|2Reα Reα
≤ 3|α−1|
Reα +· · ·+
3|α−1|
Reα =
3n|α−1|
Reα .
(21)
But|α−1| ≤ Reα
3n so from (7) we obtain that
1− |z|2Reα Reα
zf′′(z)
f′(z) ≤1,
(22)
for all z ∈ U and the Theorem 2 implies that the function Gα,n is in the class
S.
Corollary 1. Let g∈Asatisfy (2) andαa complex number such that
|α−1| ≤ Reα
3k , k∈N
∗
If|g(z)| ≤1,∀z∈U, then the function
Gkα(z) =
(k(α−1) + 1) z Z
0
gk(α−1)(t) dt
1
k(α−1)+1 (24)
is univalent.
Proof. We consider the functions
Gk α(z) =
(k(α−1) + 1) z Z
0
tk(α−1)
g(t) t
k(α−1)
dt
1
k(α−1)+1 (25)
and
f(z) =
z Z
0
g(t) t
k(α−1)
dt. (26)
The functionf is regular inU.From (26) we obtain
f′(z) =
g(z) z
k(α−1)
and
f′′(z) =k(α−1)
g(z) z
k(α−1)−1
zg′(z)−g(z)
z2 .
Next we have 1− |z|2Reα
Reα
zf′′(z)
f′(z)
≤1− |z| 2Reα
Reα k|α−1|
zg′(z)
g(z)
+ 1
,
∀z∈U. (27)
Applying the Schwartz Lemma and using (27) we obtain
1− |z|2Reα Reα
zf′′(z) f′(z)
≤ |α−1|1− |z| 2Reα
Reα
z2g′(z) g2(z) −1
+ 2
. (28)
Sincegsatisfies conditions (2) then from (28) and (23) we obtain
1− |z|2Reα Reα
zf′′(z) f′(z)
≤ 3k|α−1| Reα
1− |z|2Reα≤ 3k|α−1| Reα ≤1. (29)
Now Theorem 2 and (29) imply thatGk
α∈S.
Corollary 2. Let f, g ∈ A, satisfy (1) and α the complex number with the property
If|f(z)| ≤1,∀z∈U and|g(z)| ≤1,∀z∈U, then the function
Gα(z) =
(2α−1) z Z
0
f(α−1)(t)g(α−1)(t) dt
1 2α−1 (31)
is univalent.
Proof. In Theorem 1 we setn= 2, g1=f,g2=g.
Remark 1. Theorem 1 is a generalization of Theorem 3.
Remark 2. From Corollary 1, fork= 1, we obtain Theorem 3.
References
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Inc. 1975).
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286.
3. Ozaki S. and Nunokawa M.,The Schwartzian derivative and univalent functions, Proc. Amer.
Math. Soc.33(2) (1972), 392–394.
4. Pascu N. N.,On univalent criterion II, Itinerant seminar on functional equations
approxima-tion and convexity, Cluj-Napoca, Preprint nr.6, (1985), 153–154.
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Commem-orative Session Cimion Stoilow, Brasov (1987), 43–48.
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Mathe-matica,XXXIII(2000), 51–54.
D. Breaz, Department of Mathematics, “1 Decembrie 1918” University, Alba Iulia, Romania, e-mail:dbreaz@uab.ro
